Maths · Scalar and vector products
Equation of the plane containing the +\boldsymbol{t}(\overline{\boldsymbol{i}}+\
Equation of the plane containing the \( \operatorname{lines} \overline{\boldsymbol{r}}=(\overline{\boldsymbol{i}}-\boldsymbol{2} \overline{\boldsymbol{j}}+\overline{\boldsymbol{k}})+\boldsymbol{t}(\overline{\boldsymbol{i}}+\mathbf{2} \overline{\boldsymbol{j}}-\overline{\boldsymbol{k}}) \) \( \boldsymbol{\boldsymbol { r }}=(\overline{\boldsymbol{i}}+\mathbf{2} \overline{\boldsymbol{j}}-\overline{\boldsymbol{k}})+\boldsymbol{s}(\overline{\boldsymbol{i}}+\overline{\boldsymbol{j}}+\mathbf{3} \overline{\boldsymbol{k}}) \) is
- A. \( \bar{r}(7 \bar{i}-4 \bar{j}-\bar{k})=14 \)
- B. \( \bar{r}(\bar{i}+2 \bar{j}-\bar{k})=10 \)
- C. \( \bar{r}(\bar{i}+\bar{j}+3 \bar{k})=20 \)
- D. \( \bar{r}(\bar{i}-2 \bar{j}+\bar{k})=27 \)
Step-by-step solution
The direction vectors of the given lines are d1 = (1, 2, -1) and d2 = (1, 1, 3). Their cross product d1 × d2 = (7, -4, -1) gives a normal vector to the plane containing the lines. Using the point on the first line (1, -2, 1), the plane equation is r·(7, -4, -1) = (1, -2, 1)·(7, -4, -1) = 7 + 8 - 1 = 14.
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